1. The Classification of Stellar Spectra
Atoms/Ions in stellar atmospheres are excited and ionized primarily by collisions with atoms/ions/electrons (along with a small contribution from the absorption of photons). The temperature dependence of stellar absorption lines reflect how such collisions affect the excitation and ionization of atoms/ions.

2. Ionization of Atoms/Ions in Stellar Atmospheres
Atoms/Ions in stellar atmospheres are ionized primarily by collisions between atoms/ions/electrons.
To ionize an atom/ion from a lower to higher excitation state, the colliding atom/ion/electron must have a kinetic energy equal to or greater than the corresponding ionization energy of the other atom/ion.
To understand how the strength of stellar photospheric absorption lines depend on temperature, we not only have to understand how the distribution of atoms/ions in different excited states depend on temperature but also how their distribution in different ionization states depend on temperature.

3. Learning Objectives
Collisional Ionization of Atoms/Ions Hydrogen Helium
Distribution of Ionization States Partition Function Saha equation Degree of ionization in stellar atmospheres
Strength of Stellar Absorption Lines Combining collisional excitation and ionization Relative line strengths of different elements
Relative Abundance of Different Elements

4. Learning Objectives
Collisional Ionization of Atoms/Ions Hydrogen Helium
Distribution of Ionization States Partition Function Saha equation Degree of ionization in stellar atmospheres
Strength of Stellar Absorption Lines Combining collisional excitation and ionization Relative line strengths of different elements
Relative Abundance of Different Elements

5. Collisional Ionization of Atoms/Ions
Consider two hydrogen atoms in their ground state; i.e., with their individual electrons in the n = 1 quantum state.
In a collision between these two atoms, a part of their kinetic energy can be transferred into ionizing one or both atoms; i.e., cause their individual electrons to make a transition from the n = 1 to a free state.
Consider the solar photosphere, which is at a temperature of 5778 K: - vmp = m/s, corresponding to a kinetic energy of 0.50 eV - ‹v› = m/s , corresponding to a kinetic energy of 0.64 eV vrms = m/s, corresponding to an average kinetic energy of 0.75 eV
An energy of 13.6 eV (4.3 × vrms) is required to ionize a hydrogen atom from the ground state. Thus, in the solar photosphere, only collisions between hydrogen atoms at the tail of the Maxwell-Boltzmann velocity distribution can ionize hydrogen atoms from the ground state. (Recall that an energy of 3.7 × vrms is required to excite a hydrogen atom from the ground to its first excited state.)

6. Collisional Ionization of Atoms/Ions
Because higher excited states are less populated than lower excited states, one might think that the ionized state is even less populated than excited states. Yet the fact that hydrogen Balmer lines become weaker from A0 → B → O stars imply otherwise. How can the ionized be more populated than excited states?
An electron has to “find” an ion.
An electron has to come close enough to the ion to be influenced by its positive charge compared with surrounding ions. But ions and electrons are also moving, so an electron may only be under the influence of a single ion for a short time.

7. Collisional Ionization of Atoms/Ions
Recall that Canon placed O stars, which have visible lines of ionized helium (He II), …
Note that this photograph is a negative, so that bright lines correspond to absorption lines.

8. Collisional Ionization of Atoms/Ions
Recall that Canon placed O stars, which have visible lines of ionized helium (He II), before B stars, which do not have He II lines but have the most intense neutral helium (He I) lines, and both O and B before A stars despite having weaker H lines.
Note that this photograph is a negative, so that bright lines correspond to absorption lines.

9. Collisional Ionization of Atoms/Ions
Energy diagram of helium and permitted transitions to the n = 1 and n = 2 states. (In this rendering of the energy diagram, the ground state of parahelium is defined to have an energy of 0 eV).
Parahelium
Orthohelium

10. Collisional Ionization of Atoms/Ions
Cannon must have reasoned that it requires more energy to ionize He than H.
Around 1895, Carl Runge and Louis Paschen discovered that helium comprised
a mixture of two gases, named parahelium and orthohelium, that produced different spectral lines. Notice that the lines of both parahelium and orthohelium are weaker in O than in B stars, implying that both species are more highly ionized in O than in B stars. Ionization of parahelium requires more energy than hydrogen.

11. Learning Objectives
Collisional Ionization of Atoms/Ions Hydrogen Helium
Distribution of Ionization States Partition Function Saha equation Degree of ionization in stellar atmospheres
Strength of Stellar Absorption Lines Combining collisional excitation and ionization Relative line strengths of different elements
Relative Abundance of Different Elements

12. Distribution of Ionization States
Atoms/Ions of a gas can gain energy during a collision in the form of - kinetic energy excitation of an electron to a higher energy level removal of an electron to a free state (ionization)
Atoms/Ions of a gas also can lose energy during a collision in the form of - kinetic energy de-excitation of an electron to a lower energy level (energy transferred as kinetic energy to another atom)
The collision between an ion and electron can also result in recombination.
In a collection of atoms/ions excited and ionized by collisions, the distribution of excited and ionization states depends on the distribution in kinetic energies of the atoms. In stellar atmospheres, the distribution in speeds of the impacting atoms − given by the Maxwell-Boltzmann velocity distribution − produces a definite distribution of excited and ionization states.

13. The Partition Function
Recall that the ratio of the number of atoms ni with a higher energy Ei to the number of atoms nj with a lower energy Ej in different states of excitation is given by the Boltzmann equation
Expressed in terms of nj = n1, the number of atoms in the ground state,
The total number of atoms in all states is therefore
Eij = Ei - Ej
Ei = Ei – E1
Partition Function, Z

14. The Partition Function
By defining the partition function as
the number of atoms in each excited state can be written in terms of the partition function
The above equation tells us that the number of atoms in a given excited state i depends on the distribution of atoms in different excited states as encoded in the partition function Z. What is the reason for this dependence?
Ei = Ei – E1

15. The Partition Function
By defining the partition function as
the number of atoms in each excited state can be written in terms of the partition function
The partition function encodes the statistical properties (in this case, distribution of excited states) of a system in thermal (more accurately, thermodynamic) equilibrium. As an illustration, if the hydrogen atom had only one possible electron orbital (n = 1), what would the partition function be?
Ei = Ei – E1

16. The Partition Function
The partition function (returning to the notation used in the textbook)
Let us compute the partition function for hydrogen in the solar photosphere, where the temperature is 5778 K. What would you expect the partition function to be in this case?

17. Collisional Ionization of Atoms
The ratio of the number of atoms in ionization stage i (i = 0, 1, 2, … depending on the element) to the number of atoms in stage i is given by the Saha equation
E.g., in the case of hydrogen NI is neutral hydrogen, and NII is ionized hydrogen. This equation was first derived by the Indian astrophysicist Meghnad Saha in 1920.
The Saha equation is derived by equating the ionization and recombination rate for the i and i+1 ionization stage.
ionization energy from ground state
Meghnad Saha,

18. Collisional Ionization of Atoms
The ratio of the number of atoms in ionization stage i (i = 0, 1, 2, … depending on the element) to the number of atoms in stage i is given by the Saha equation
E.g., in the case of hydrogen NI is neutral hydrogen, and NII is ionized hydrogen. This equation was first derived by the Indian astrophysicist Meghnad Saha in 1920.
Notice that the number fraction of ions to neutrals depends on ionization energy; as χ1 ↑, Ni+1/Ni temperature; as T ↑, Ni+1/Ni ne; as ne ↑, Ni+1/Ni the partition function for the atoms/ions; why?
ionization energy from ground state
Meghnad Saha,
As T increases, Z also increases. The temperature dependence of ionization is a function of both Z and T, with the T term dominating.

19. Collisional Ionization of Atoms
The ratio of the number of atoms in ionization stage i (i = 0, 1, 2, … depending on the element) to the number of atoms in stage i is given by the Saha equation
E.g., in the case of hydrogen NI is neutral hydrogen, and NII is ionized hydrogen. This equation was first derived by the Indian astrophysicist Meghnad Saha in 1920.
The Saha equation is often expressed in terms of electron pressure rather than electron density through the ideal gas equation Pe = nekT so that
ionization energy from ground state
Meghnad Saha,

20. Degree of Ionization in Stellar Atmospheres
Let us compute the degree of ionization in stellar photospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2.
In this temperature range, the partition function ZI = 2 (most of the hydrogen atoms are in the ground state, and neglecting high order terms). A hydrogen ion is just a proton, and has no energy levels or degeneracy so that ZII = 1.
Inserting into the Saha equation, we can solve for NII/NI; i.e., the ratio of ionized to neutral hydrogen.

21. Degree of Ionization in Stellar Atmospheres
Let us compute the degree of ionization in stellar photospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2.
In this temperature range, the partition function ZI = 2 (most of the hydrogen atoms are in the ground state, and neglecting high order terms). A hydrogen ion is just a proton, and has no energy levels or degeneracy so that ZII = 1.
Inserting into the Saha equation, we can solve for NII/NI; i.e., the ratio of ionized to neutral hydrogen.
To compute the fraction of hydrogen ions to the total number of hydrogen (atoms+ions), we write

22. Degree of Ionization in Stellar Atmospheres
In stellar photospheres, hydrogen atoms are ionized over a narrow range of temperatures around 10,000 K.

23. Degree of Ionization in Stellar Atmospheres
In stellar photospheres, hydrogen atoms are ionized over a narrow range of temperatures around 10,000 K.
At a gas temperature of 10,000 K: vmp = m/s, corresponding to a kinetic energy of 0.86 eV - ‹v› = m/s , corresponding to a kinetic energy of 1.10 eV vrms = m/s, corresponding to an average kinetic energy of 1.29 eV
An energy of 13.6 eV (3.2 × vrms) is required to ionize a hydrogen atom from the ground state. Thus, collisions that result in ionization involve hydrogen atoms at the tail of the Maxwell-Boltzmann velocity distribution, as well as between hydrogen atoms with lower speeds and excited hydrogen atoms.

24. Degree of Ionization in Stellar Atmospheres
Notice that the fraction of ionized hydrogen rises significantly just as the fraction of hydrogen atoms in the first excited state starts to rise. Why?
Collisional Excitation
Collisional Ionization

25. Learning Objectives
Collisional Ionization of Atoms/Ions Hydrogen Helium
Distribution of Ionization States Partition Function Saha equation Degree of ionization in stellar atmospheres
Strength of Stellar Absorption Lines Combining collisional excitation and ionization Relative line strengths of different elements
Relative Abundance of Different Elements

26. Combining Collisional Excitation and Collisional Ionization
The strength of absorption lines in stellar spectra – i.e., the number of atoms/ions in the ground or excited state involved in the transition that produces a particular absorption line – depends on the effects of both collisional excitation and collisional ionization.
Collisional Excitation
Collisional Ionization

27. Combining Collisional Excitation and Collisional Ionization
The strength of absorption lines in stellar spectra – i.e., the number of atoms/ions in the ground or excited state involved in the transition that produces a particular absorption line – depends on the effects of both collisional excitation and collisional ionization.
E.g., consider the Balmer absorption lines of hydrogen in stellar spectra, produced by hydrogen atoms in the first excited state. As the temperature increases, the fraction of hydrogen atoms in the first excited state to that in the ground state increases, and so the strength of Balmer absorption lines also increases. As the temperature increases, however, an increasing fraction of hydrogen atoms become ionized, leaving fewer neutral hydrogen atoms available to produce (Balmer) absorption lines. Thus, the combination of both collisional excitation and collisional ionization determines how the strength of absorption lines depend on gas temperature, and therefore how the strength of absorption lines in stellar spectra depend on stellar effective temperature.

28. Combining Collisional Excitation and Collisional Ionization
Consider the Balmer absorption lines of hydrogen in stellar atmospheres, produced by the absorption of photons by hydrogen atoms in the first excited state (i.e., electron in the n = 2 orbital).
The strength of the Balmer lines depends on N2/Ntotal, the fraction of all hydrogen particles that are in the first excited state.
Because nearly all hydrogen atoms are in the ground state and most of the remainder in the first excited state so that N1 + N2 ≅ NI, we can write
Total number of neutral hydrogen atoms
Total number of hydrogen atoms and ions (NI + NII)
Compute from Boltzmann equation
Compute from Saha equation

29. Combining Collisional Excitation and Collisional Ionization
Plot of N2/Ntotal in stellar atmospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2.
Collisional Excitation
Collisional Ionization

30. Combining Collisional Excitation and Collisional Ionization
Plot of N2/Ntotal in stellar atmospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2. Do you now understand why the strength of Balmer absorption lines reaches a maximum at spectral type A0?

31. Relative Line Strengths of Different Elements
In the spectra of the Sun and other G-type stars, the absorption lines of calcium are far stronger than the Balmer lines of hydrogen. Why is this the case, given that the cosmic abundance of calcium is far smaller than that of hydrogen?

32. Relative Line Strengths of Different Elements
Let us first compute the fraction of ionized to neutral hydrogen. The electron pressure in the solar photosphere is 1.5 N m-2.
From the Saha equation, the ratio of ionized to neutral hydrogen
Thus, there is only one hydrogen ion (H II) for every 13,000 hydrogen atoms (H I). Almost all the hydrogen in the solar photosphere is neutral.

33. Relative Line Strengths of Different Elements
This is in agreement with the previous plot (redisplayed below) showing the degree of ionization in stellar photospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2

34. Relative Line Strengths of Different Elements
Let us now compute the fraction of hydrogen atoms in the first excited state (capable of producing the Balmer absorption lines) to the ground state.
From the Boltzmann equation, the ratio of hydrogen atoms in the first excited state to the ground state
Thus, there is only one hydrogen atom in the first exited state to every 198,000,000 hydrogen atoms in the ground state. Almost all the hydrogen atoms in the solar photosphere are in the ground state.

35. Distribution of Excited States
This is in agreement with the earlier plot (redisplayed below) of the relative number of hydrogen atoms in the first exited state to the total number of hydrogen atoms in the ground and first excited states.

36. = 1/200,000,000 Relative Line Strengths of Different Elements
Combining the Botzmann and Saha equations, the fraction of hydrogen atoms in the first excited state (capable of producing the Balmer absorption lines) to the total number of hydrogen (atoms+ions) (employing N1 + N2 ≅ NI) is
Thus, there is only one hydrogen atom in the first excited state (capable of producing Balmer absorption) to every 200,000,000 hydrogen particles (mostly neutral and in the ground state) in the solar photosphere.
= 1/200,000,000

37. Distribution of Excited States
This is in agreement with the earlier plot (redisplayed below) of the relative number of hydrogen atoms in the first exited state compared to the total number of hydrogen (atoms and ions).

38. Relative Line Strengths of Different Elements
Let us compute the fraction of singly-ionized calcium atoms (Ca II) in the ground state (capable of producing the Ca II H and K absorption lines) to the total number of calcium atoms in the solar photosphere.
The ionization energy of neutral calcium (to single-ionized calcium) is 6.11 eV, only about half of the 13.6 eV ionization energy of hydrogen.
From the Saha equation, the ratio of singly-ionized to neutral calcium is (ZI = 1.32 and ZII = 2.30 for calcium)
Practically all of the calcium atoms in the solar photosphere are singly-ionized (in the form of Ca II), with only one of every 918 calcium atoms remaining neutral (in the form of Ca I). Note: the ionization energy of Ca II to Ca III is 11.9 eV, and so there are very few doubly-ionized calcium atoms (Ca III).

39. Relative Line Strengths of Different Elements
Let us now compute the fraction of singly-ionized calcium atoms (Ca II) in the ground state (capable of producing the Ca II H and K absorption lines) to the total number of calcium atoms.
The first excited state of Ca II is E2 - E1 = 3.12 eV above the ground state.
From the Boltzmann equation, the ratio of Ca II atoms in the first excited state to the ground state
Thus, there is only one Ca II atom in the first exited state to every 264 Ca II atoms in the ground state. Almost all of the Ca II atoms in the solar photosphere are in the ground state.

40. [ ]Ca II Relative Line Strengths of Different Elements
Combining the Boltzmann and Saha equations, the fraction of singly-ionized calcium atoms (Ca II) in the ground state (capable of producing the Ca II H and K absorption lines) to the total number of calcium (atoms and ions) is
Thus, 995 out of 1000 calcium atoms are singly ionized and in the ground state. Almost all calcium atoms in the solar photosphere are singly-ionized and in the ground state. [
]Ca II

41. Relative Line Strengths of Different Elements
There are 500,000 hydrogen atoms to every calcium atom in the solar photosphere. However, the fraction of hydrogen atoms in the first excited state to all hydrogen is only 1/200,000,000. By comparison, virtually all calcium atoms in the solar photosphere are singly-ionized and in the ground state. Thus, for every hydrogen atom in the first excited state capable of producing Balmer absorption line, there are 400 singly-ionized calcium atoms in the ground state capable of producing the Ca II H and K lines.
This is the reason why the Ca II Ha and K lines in the spectrum of the Sun is so much stronger than the Balmer lines of hydrogen.

42. Relative Line Strengths of Different Elements
Plot of the strength of various spectral lines in stellar atmospheres as a function of stellar effective temperature as calculated from the Boltzmann and Saha equations.

43. Learning Objectives
Collisional Ionization of Atoms/Ions Hydrogen Helium
Distribution of Ionization States Partition Function Saha equation Degree of ionization in stellar atmospheres
Strength of Stellar Absorption Lines Combining collisional excitation and ionization Relative line strengths of different elements
Relative Abundance of Different Elements

44. Relative Abundance of Different Elements
How do we know the cosmic abundance of the elements?
Until the mid-1920’s, the accepted wisdom was that the relative abundance of different elements in the Sun was the same as that on the Earth, ~65% iron and ~35% hydrogen.
Solar Spectrum

45. Relative Abundance of Different Elements
If we know the relative abundances of different elements and how their line strengths depend on temperature (through the Boltzmann and Saha equations), we can compute the relative line strengths of different elements.
Similarly, if we know the relative line strengths of different elements (from stellar spectra) and how their line strengths depend on temperature (through the Boltzmann and Saha equations), we can compute the relative abundances of different elements.
The first person to do so was Cecilia Payne, who (using the Boltzmann and Saha equations) in 1925 calculated the abundance of 18 elements in the stellar photospheres in her PhD thesis. She showed for the first time that hydrogen is the most abundant element in the Universe.
Cecilia Payne,